Chapter 8 · Organic Chemistry – Some Basic Principles and Techniques

8.3 Reaction Mechanism Basics: Electron Displacement and Intermediates

Organic reactions happen because electrons move: bonds break evenly or unevenly, short-lived carbocations, carbanions and free radicals appear, and electron-rich reagents meet electron-poor sites. This section introduces curved-arrow notation, electrophiles and nucleophiles, the inductive, electromeric and resonance effects, hyperconjugation, and the four main types of organic reaction.

Leave a mixture of methane and chlorine in the dark and nothing happens. Shine sunlight on it and it reacts explosively, giving chloromethane and hydrogen chloride. Bromoethane, on the other hand, reacts with aqueous potassium hydroxide in the dark, quietly, giving ethanol. Two substitution reactions that look alike on paper go by completely different routes: the first through neutral fragments with unpaired electrons, the second through an attack by a negative ion on a slightly positive carbon atom.

The step-by-step story of which bonds break, which form and where the electrons go is called the reaction mechanism. A few ideas explain most organic mechanisms: how a covalent bond can break, which reactive intermediates result and how stable they are, which reagents seek electrons and which offer them, and how the electrons of a molecule are pulled or pushed by the atoms and groups it contains. These ideas — met again in every chapter of Class 12 organic chemistry — are the subject of this section.

Breaking Bonds: Fission, Intermediates and Reagents

📘 Definition 8.8 — Homolytic and heterolytic fission
A covalent bond $\mathrm{A{-}B}$ (a shared pair of electrons) can break in two ways:
  1. Homolytic fission (homolysis): each atom takes one electron of the pair, giving two neutral species with an unpaired electron, called free radicals: $\mathrm{A{-}B\rightarrow A^{\bullet}+{}^{\bullet}B}$. It is favoured by heat, light ($h\nu$) or peroxides, by non-polar bonds and by the gas phase or non-polar solvents.
  2. Heterolytic fission (heterolysis): the more electronegative atom takes both electrons, giving ions: $\mathrm{A{-}B\rightarrow A^{+}+B^{-}}$ when B is more electronegative. It is favoured by polar bonds and polar solvents, which stabilise the ions.
Curved arrows show the movement of electrons, always from where the electrons are (a bond or a lone pair) to where they go. A full-headed arrow moves an electron pair; a half-headed ("fish-hook") arrow moves a single electron.
📘 Definition 8.9 — Carbocations, carbanions and free radicals
These short-lived reactive intermediates are named after the carbon atom that carries the charge or the odd electron, which is called primary (1°), secondary (2°) or tertiary (3°) if it is bonded to one, two or three other carbon atoms.
  • A carbocation has a carbon with only six valence electrons and a positive charge, e.g. $\mathrm{CH_{3}^{+}}$, $\mathrm{(CH_{3})_{3}C^{+}}$. The carbon is $sp^{2}$: trigonal planar, with an empty $p$ orbital.
  • A carbanion has a carbon with eight valence electrons, one pair unshared, and a negative charge, e.g. $\mathrm{CH_{3}^{-}}$. The carbon is $sp^{3}$: pyramidal, like $\mathrm{NH_{3}}$, with which $\mathrm{CH_{3}^{-}}$ is isoelectronic.
  • A free radical has a carbon with seven valence electrons, one of them unpaired, and no charge, e.g. $\mathrm{CH_{3}^{\bullet}}$. Alkyl radicals are (nearly) planar, $sp^{2}$, with the odd electron in a $p$ orbital; they are paramagnetic.
📐 Rule 8.8 — Stability of reactive intermediates
Alkyl groups release electrons towards the carbon they are attached to (+I effect and hyperconjugation, explained below). An electron-poor centre is stabilised by this, an electron-rich centre destabilised: $$\text{carbocations, free radicals:}\quad 3^{\circ}\gt2^{\circ}\gt1^{\circ}\gt\mathrm{CH_{3}}$$ $$\text{carbanions:}\quad \mathrm{CH_{3}^{-}}\gt1^{\circ}\gt2^{\circ}\gt3^{\circ}$$ For example $\mathrm{(CH_{3})_{3}C^{+}}\gt\mathrm{(CH_{3})_{2}CH^{+}}\gt\mathrm{CH_{3}CH_{2}^{+}}\gt\mathrm{CH_{3}^{+}}$. Intermediates next to a double bond or a benzene ring (allyl $\mathrm{CH_{2}{=}CH{-}CH_{2}^{+}}$, benzyl $\mathrm{C_{6}H_{5}CH_{2}^{+}}$, and the corresponding radicals and anions) are further stabilised because the charge or odd electron is spread by resonance.
📘 Definition 8.10 — Electrophiles and nucleophiles
In most reactions an attacking reagent meets an organic substrate.
  • A nucleophile ("nucleus-loving", $\mathrm{Nu^{:}}$) is electron-rich and attacks an electron-poor centre by donating an electron pair: anions such as $\mathrm{OH^{-}}$, $\mathrm{CN^{-}}$, $\mathrm{Cl^{-}}$, $\mathrm{RO^{-}}$, carbanions; neutral molecules with lone pairs such as $\mathrm{H_{2}O}$, $\mathrm{NH_{3}}$, $\mathrm{R{-}NH_{2}}$, $\mathrm{R{-}OH}$.
  • An electrophile ("electron-loving", $\mathrm{E^{+}}$) is electron-poor and accepts an electron pair: $\mathrm{H^{+}}$, $\mathrm{NO_{2}^{+}}$, $\mathrm{Cl^{+}}$, carbocations; neutral molecules with an incomplete octet or a strongly electron-poor atom, such as $\mathrm{BF_{3}}$, $\mathrm{AlCl_{3}}$, $\mathrm{SO_{3}}$, and the carbon atom of a carbonyl group.
A curved arrow in a mechanism therefore usually starts on the nucleophile (its lone pair or π bond) and ends on the electrophile.
✏️ Example 1 — Following a bond fission
For each fission, name the type, give the products and say what reactive intermediate (if any) is formed, with the hybridisation of its carbon: (a) $\mathrm{CH_{3}{-}CH_{3}}$ heated strongly; (b) $\mathrm{(CH_{3})_{3}C{-}Cl}$ dissolved in water; (c) the $\mathrm{C{-}Li}$ bond of $\mathrm{CH_{3}Li}$ breaking to ions; (d) $\mathrm{H{-}Br}$ dissolving in water; (e) $\mathrm{Br{-}Br}$ in sunlight.
(a) Non-polar bond, heat: homolysis, $\mathrm{CH_{3}{-}CH_{3}\rightarrow2\,CH_{3}^{\bullet}}$ — two methyl free radicals, carbon $sp^{2}$ (planar, 7 valence electrons).
(b) Polar bond, polar solvent: heterolysis; Cl ($\chi=3.0$) takes the pair from C ($\chi=2.5$): $\mathrm{(CH_{3})_{3}C^{+}+Cl^{-}}$ — a tertiary carbocation, $sp^{2}$.
(c) Here carbon ($\chi=2.5$) is more electronegative than lithium ($\chi=1.0$), so carbon keeps the pair: $\mathrm{CH_{3}^{-}+Li^{+}}$ — a carbanion, $sp^{3}$, pyramidal.
(d), (e) HBr: heterolysis, $\mathrm{H^{+}+Br^{-}}$ (no carbon intermediate; the ions are hydrated). $\mathrm{Br_{2}}$ in light: homolysis, $\mathrm{2\,Br^{\bullet}}$ — bromine free radicals.
Check Charges balance in every case: neutral reactant → two neutral radicals, or one cation + one anion ✓.
✏️ Example 2 — Electrophile or nucleophile?
Classify each species and give a reason: $\mathrm{H_{2}O}$, $\mathrm{BF_{3}}$, $\mathrm{CN^{-}}$, $\mathrm{NO_{2}^{+}}$, $\mathrm{NH_{3}}$, $\mathrm{(CH_{3})_{3}C^{+}}$, $\mathrm{AlCl_{3}}$, $\mathrm{CH_{3}O^{-}}$, the carbon atom of $\mathrm{HCHO}$, $\mathrm{(C_{2}H_{5})_{3}N}$.
Nucleophiles $\mathrm{H_{2}O}$ and $\mathrm{NH_{3}}$ (lone pairs on O, N); $\mathrm{CN^{-}}$ and $\mathrm{CH_{3}O^{-}}$ (anions with lone pairs); $\mathrm{(C_{2}H_{5})_{3}N}$ (lone pair on N).
Electrophiles $\mathrm{NO_{2}^{+}}$ and $\mathrm{(CH_{3})_{3}C^{+}}$ (positive, electron-poor); $\mathrm{BF_{3}}$ and $\mathrm{AlCl_{3}}$ (only six electrons round B and Al — they accept a pair); the carbonyl carbon of $\mathrm{HCHO}$ (O pulls the $\mathrm{C{=}O}$ electrons away, leaving C with $\delta+$).
Check Every nucleophile has at least one lone pair or a negative charge; every electrophile has an empty orbital or a strongly electron-poor atom ✓.
Interactive 1 — Homolytic vs Heterolytic Fission

Choose a bond and a way of breaking it, then press ▶ Play. The two bonding electrons (black dots) travel along the curved arrows: half-headed arrows carry one electron each, a full-headed arrow carries the pair to the more electronegative atom (Pauling electronegativities shown). The lab names the fragments, counts the valence electrons on carbon and says whether this fission is realistic and under what conditions.

products—
valence e⁻ on carbon—
intermediate—
favoured by—

Electron Displacement: Inductive and Electromeric Effects

Where will a nucleophile or an electrophile attack, and how stable is the intermediate formed? Both depend on how the electrons of a molecule are displaced by the atoms and groups it contains. Some displacements are permanent (present in the molecule at all times); others appear only when a reagent approaches.

📘 Definition 8.11 — Inductive effect
The inductive effect is the permanent displacement of the σ-bond electrons along a carbon chain caused by an atom or group of different electronegativity. In $\mathrm{CH_{3}{-}CH_{2}{-}CH_{2}{-}Cl}$ chlorine pulls the $\mathrm{C1{-}Cl}$ electrons towards itself; C1 becomes $\delta+$, and C1 in turn pulls slightly on the $\mathrm{C2{-}C1}$ electrons, so C2 becomes $\delta\delta+$, C3 $\delta\delta\delta+$. The effect fades quickly and is negligible beyond about three bonds.
  • −I groups (electron-withdrawing): $\mathrm{{-}NO_{2}}$, $\mathrm{{-}CN}$, $\mathrm{{-}COOH}$, $\mathrm{{-}CHO}$, the halogens ($\mathrm{F\gt Cl\gt Br\gt I}$), $\mathrm{{-}OH}$, $\mathrm{{-}OR}$, $\mathrm{{-}C_{6}H_{5}}$.
  • +I groups (electron-releasing): alkyl groups, $\mathrm{(CH_{3})_{3}C{-}\gt(CH_{3})_{2}CH{-}\gt CH_{3}CH_{2}{-}\gt CH_{3}{-}}$, and negatively charged groups such as $\mathrm{{-}O^{-}}$, $\mathrm{{-}COO^{-}}$.

The inductive effect explains why chlorine makes ethanoic acid stronger. Ionisation gives $\mathrm{RCOO^{-}}$; a −I group next to the carboxylate spreads its negative charge and stabilises it, so the acid gives up $\mathrm{H^{+}}$ more readily. A smaller $\mathrm{p}K_{a}$ means a stronger acid (Chapter 6):

Acid$\mathrm{CH_{3}COOH}$$\mathrm{ClCH_{2}COOH}$$\mathrm{Cl_{2}CHCOOH}$$\mathrm{Cl_{3}CCOOH}$$\mathrm{FCH_{2}COOH}$$\mathrm{BrCH_{2}COOH}$$\mathrm{ICH_{2}COOH}$
$\mathrm{p}K_{a}$ (approx.)4.762.861.290.652.592.903.18
📘 Definition 8.12 — Electromeric effect
The electromeric effect (E effect) is a temporary effect: when an attacking reagent approaches a multiple bond, the π electrons are transferred completely to one of the two atoms. In the +E effect the π electrons move towards the atom to which the reagent becomes attached (e.g. $\mathrm{H^{+}}$ adding to $\mathrm{C{=}C}$); in the −E effect they move away from the atom attacked (e.g. $\mathrm{CN^{-}}$ attacking the carbon of $\mathrm{C{=}O}$, the electrons going to O). The effect disappears as soon as the reagent is removed. When inductive and electromeric effects oppose each other, the electromeric effect usually wins.
✏️ Example 3 — Using the inductive effect
(a) Arrange $\mathrm{CH_{3}COOH}$, $\mathrm{ClCH_{2}COOH}$, $\mathrm{Cl_{2}CHCOOH}$ and $\mathrm{Cl_{3}CCOOH}$ in order of increasing acid strength. (b) Arrange the four haloethanoic acids $\mathrm{XCH_{2}COOH}$ ($\mathrm{X}=\mathrm{F,Cl,Br,I}$) in order of decreasing acid strength. (c) Arrange 2-, 3- and 4-chlorobutanoic acid in order of decreasing acid strength. (d) Show the partial charges along the chain of 1-chloropropane. (e) Which alkyl group has the strongest +I effect: $\mathrm{CH_{3}{-}}$, $\mathrm{C_{2}H_{5}{-}}$, $\mathrm{(CH_{3})_{2}CH{-}}$ or $\mathrm{(CH_{3})_{3}C{-}}$?
(a) Each extra Cl adds a −I pull: $\mathrm{CH_{3}COOH}\lt\mathrm{ClCH_{2}COOH}\lt\mathrm{Cl_{2}CHCOOH}\lt\mathrm{Cl_{3}CCOOH}$ ($\mathrm{p}K_{a}$ 4.76 → 2.86 → 1.29 → 0.65).
(b) −I follows electronegativity: $\mathrm{F\gt Cl\gt Br\gt I}$, so $\mathrm{FCH_{2}COOH\gt ClCH_{2}COOH\gt BrCH_{2}COOH\gt ICH_{2}COOH}$ ($\mathrm{p}K_{a}$ 2.59, 2.86, 2.90, 3.18).
(c) The effect fades with distance from $\mathrm{COOH}$: 2-chloro $\gt$ 3-chloro $\gt$ 4-chloro ($\mathrm{p}K_{a}$ about 2.9, 4.1 and 4.5; butanoic acid itself 4.8).
(d) $\mathrm{\overset{\delta\delta\delta+}{C}H_{3}{-}\overset{\delta\delta+}{C}H_{2}{-}\overset{\delta+}{C}H_{2}{-}\overset{\delta-}{Cl}}$: the positive charge decreases away from Cl.
(e) $\mathrm{(CH_{3})_{3}C{-}}$: each methyl pushes electron density towards the attached carbon, and tert-butyl has three of them.

Resonance and Hyperconjugation

Some molecules and ions cannot be described by a single Lewis structure (Section 4.1). In the ethanoate ion, $\mathrm{CH_{3}COO^{-}}$, both carbon–oxygen bonds are the same length (about $126\ \mathrm{pm}$, between $\mathrm{C{-}O}$ at $143\ \mathrm{pm}$ and $\mathrm{C{=}O}$ at $121\ \mathrm{pm}$), although any one Lewis structure shows one single and one double bond.

📘 Definition 8.13 — Resonance and the resonance (mesomeric) effect
When a structure can be written in two or more ways that differ only in the arrangement of π electrons and lone pairs, the real molecule is a resonance hybrid of these contributing (canonical) structures, more stable than any one of them; the extra stability is the resonance energy.
The resonance effect (R effect) is the permanent polarity produced in a conjugated system (alternating single and multiple bonds, or a lone pair next to a π bond) by the interaction of a group's lone pair or π bond with it:
  • +R effect — the group releases electrons into the conjugated system: $\mathrm{{-}OH}$, $\mathrm{{-}OR}$, $\mathrm{{-}NH_{2}}$, $\mathrm{{-}NHR}$, $\mathrm{{-}NHCOCH_{3}}$, halogens. In phenol and aniline the ring gains electron density, mainly at the ortho and para positions.
  • −R effect — the group withdraws electrons from the conjugated system: $\mathrm{{-}NO_{2}}$, $\mathrm{{-}CN}$, $\mathrm{{-}CHO}$, $\mathrm{{-}COR}$, $\mathrm{{-}COOH}$, $\mathrm{{-}COOR}$. In nitrobenzene the ortho and para positions become electron-poor.
📐 Rule 8.9 — Writing and judging resonance structures
  1. All contributing structures have the same positions of the nuclei and the same number of unpaired electrons; only π electrons and lone pairs move. (Moving an H atom gives a different compound, not a resonance structure.)
  2. Every structure must be a valid Lewis structure: no carbon with more than four bonds, no second-period atom with more than an octet.
  3. The more important (more stable) contributors have: more covalent bonds; complete octets on all atoms; less separation of opposite charges; a negative charge on the more electronegative atom (and a positive charge on the less electronegative one).
  4. Equivalent structures (as in ethanoate or benzene) contribute equally and give a large resonance stabilisation.
📘 Definition 8.14 — Hyperconjugation
Hyperconjugation is the delocalisation of the σ electrons of a $\mathrm{C{-}H}$ bond on an α-carbon (the carbon next to the charged, radical or unsaturated carbon) into an adjacent empty or partly filled $p$ orbital, or into a π bond. It is often written as "no-bond resonance": $$\mathrm{H{-}CH_{2}{-}\overset{+}{C}H_{2}\ \leftrightarrow\ H^{+}\ \ CH_{2}{=}CH_{2}}$$ Each α-hydrogen gives one such structure, so the number of α-hydrogens measures the extent of hyperconjugation: $\mathrm{(CH_{3})_{3}C^{+}}$ has 9, $\mathrm{(CH_{3})_{2}CH^{+}}$ 6, $\mathrm{CH_{3}CH_{2}^{+}}$ 3 and $\mathrm{CH_{3}^{+}}$ none — the order of Rule 8.8. For the same reason alkenes with more α-hydrogens are more stable: $\mathrm{(CH_{3})_{2}C{=}C(CH_{3})_{2}}$ (12) $\gt$ $\mathrm{(CH_{3})_{2}C{=}CHCH_{3}}$ (9) $\gt$ $\mathrm{CH_{3}CH{=}CHCH_{3}}$ (6) $\gt$ $\mathrm{CH_{3}CH{=}CH_{2}}$ (3) $\gt$ $\mathrm{CH_{2}{=}CH_{2}}$ (0).
✏️ Example 4 — Resonance structures and the R effect
(a) Write the resonance structures of the allyl cation $\mathrm{CH_{2}{=}CH{-}CH_{2}^{+}}$; on which carbons is the positive charge? (b) In phenol, the lone pair of O enters the ring. Which ring positions become electron-rich, and is $\mathrm{{-}OH}$ a +R or −R group? (c) Is $\mathrm{{-}NO_{2}}$ in nitrobenzene +R or −R, and which positions become electron-poor? (d) Are $\mathrm{CH_{3}{-}CHO}$ and $\mathrm{CH_{2}{=}CH{-}OH}$ resonance structures? (e) Which contributor of $\mathrm{CH_{2}{=}CH{-}Cl}$ is more important: $\mathrm{CH_{2}{=}CH{-}Cl}$ or $\mathrm{{}^{-}CH_{2}{-}CH{=}Cl^{+}}$?
(a) $\mathrm{CH_{2}{=}CH{-}CH_{2}^{+}\leftrightarrow{}^{+}CH_{2}{-}CH{=}CH_{2}}$ (the π pair moves towards the empty $p$ orbital). The charge is shared by C1 and C3, never C2; the two structures are equivalent, so the allyl cation is well stabilised.
(b) Pushing the O lone pair into the ring gives three charge-separated structures with the negative charge at C2, C4 and C6 (ortho, para). $\mathrm{{-}OH}$ is a +R group (it is also −I, but in phenol the +R effect dominates).
(c) −R: the ring π electrons are pulled onto the oxygens of $\mathrm{NO_{2}}$, leaving positive charge at the ortho and para carbons.
(d) No — a hydrogen atom has moved from C to O. They are two different compounds (tautomers), not two pictures of one molecule (Rule 8.9, 1).
(e) $\mathrm{CH_{2}{=}CH{-}Cl}$: no charge separation, and Cl keeps its normal one bond. The charge-separated structure (positive charge on the electronegative Cl) contributes only a little — enough to give the $\mathrm{C{-}Cl}$ bond some double-bond character.
✏️ Example 5 — Counting α-hydrogens
(a) Count the α-hydrogens in $\mathrm{CH_{3}^{+}}$, $\mathrm{CH_{3}CH_{2}^{+}}$, $\mathrm{(CH_{3})_{2}CH^{+}}$, $\mathrm{(CH_{3})_{3}C^{+}}$ and $\mathrm{(CH_{3})_{2}\overset{+}{C}CH_{2}CH_{3}}$, and arrange the first four in order of stability. (b) Give two reasons why $\mathrm{(CH_{3})_{3}C^{+}}$ is more stable than $\mathrm{CH_{3}^{+}}$. (c) Arrange ethene, propene, but-2-ene and 2-methylbut-2-ene in order of increasing stability. (d) Arrange the carbanions $\mathrm{CH_{3}^{-}}$, $\mathrm{CH_{3}CH_{2}^{-}}$, $\mathrm{(CH_{3})_{2}CH^{-}}$, $\mathrm{(CH_{3})_{3}C^{-}}$ in order of stability.
(a) 0, 3, 6, 9 and $3+3+2=8$. Stability: $\mathrm{(CH_{3})_{3}C^{+}\gt(CH_{3})_{2}CH^{+}\gt CH_{3}CH_{2}^{+}\gt CH_{3}^{+}}$.
(b) Three methyl groups release electrons by the +I effect, and nine α-C–H bonds share their electrons with the empty $p$ orbital by hyperconjugation; both spread the positive charge. $\mathrm{CH_{3}^{+}}$ has neither.
(c) α-H: 0, 3, 6, 9 → ethene $\lt$ propene $\lt$ but-2-ene $\lt$ 2-methylbut-2-ene.
(d) Alkyl groups push electrons onto an already negative carbon: $\mathrm{CH_{3}^{-}\gt CH_{3}CH_{2}^{-}\gt(CH_{3})_{2}CH^{-}\gt(CH_{3})_{3}C^{-}}$ — the reverse of the carbocation order.
Interactive 2 — Electron-Displacement Lab: Stability of Intermediates

Choose the kind of intermediate and the group around the reactive carbon. Blue arrows show the +I push of each alkyl group; the glowing $\mathrm{C{-}H}$ bonds are the α-hydrogens that take part in hyperconjugation; for allyl and benzyl the lab shows where resonance spreads the charge. The ladder on the right orders methyl, 1°, 2° and 3° and marks your choice. Press ▶ Play to step through them all.

species—
α-H (hyperconjugation)—
+I alkyl groups—
charge spread over—

Types of Organic Reactions

📘 Definition 8.15 — The four main types of organic reaction
  1. Substitution — an atom or group is replaced by another: $\mathrm{CH_{3}CH_{2}Br(l)+OH^{-}(aq)\rightarrow CH_{3}CH_{2}OH(aq)+Br^{-}(aq)}$ (nucleophilic substitution); $\mathrm{CH_{4}(g)+Cl_{2}(g)\xrightarrow{h\nu}CH_{3}Cl(g)+HCl(g)}$ (free-radical substitution).
  2. Addition — two molecules combine into one, typically across a multiple bond: $\mathrm{CH_{2}{=}CH_{2}(g)+Br_{2}(l)\rightarrow CH_{2}BrCH_{2}Br(l)}$.
  3. Elimination — atoms or groups are removed from neighbouring carbons, forming a multiple bond: $\mathrm{CH_{3}CH_{2}OH(l)\xrightarrow[443\ \mathrm{K}]{\text{conc. }H_{2}SO_{4}}CH_{2}{=}CH_{2}(g)+H_{2}O(l)}$.
  4. Rearrangement — the atoms of a molecule are reorganised into an isomer: $\mathrm{CH_{3}CH_{2}CH_{2}CH_{3}(g)\xrightarrow{\text{anhyd. }AlCl_{3}/HCl}(CH_{3})_{2}CHCH_{3}(g)}$.
Substitution and addition are further described by the attacking reagent: nucleophilic, electrophilic or free-radical.
✏️ Example 6 — Classifying reactions
Classify each reaction and identify the attacking reagent as a nucleophile, an electrophile or a free radical: (a) $\mathrm{CH_{3}Cl+CN^{-}\rightarrow CH_{3}CN+Cl^{-}}$; (b) $\mathrm{CH_{2}{=}CH_{2}+HBr\rightarrow CH_{3}CH_{2}Br}$ (the first step is attack by $\mathrm{H^{+}}$); (c) $\mathrm{CH_{3}CH_{2}Br+KOH\ (alc.)\rightarrow CH_{2}{=}CH_{2}+KBr+H_{2}O}$; (d) $\mathrm{CH_{3}CHO+HCN\rightarrow CH_{3}CH(OH)CN}$ (the first step is attack by $\mathrm{CN^{-}}$ on the carbonyl carbon); (e) $\mathrm{C_{2}H_{6}+Cl_{2}\xrightarrow{h\nu}C_{2}H_{5}Cl+HCl}$; (f) butane → 2-methylpropane over $\mathrm{AlCl_{3}}$.
(a) Substitution: $\mathrm{CN^{-}}$ (nucleophile) replaces $\mathrm{Cl^{-}}$ at the $\delta+$ carbon — nucleophilic substitution.
(b) Addition across $\mathrm{C{=}C}$ started by the electrophile $\mathrm{H^{+}}$ (+E effect of the π bond) — electrophilic addition.
(c) Elimination of HBr from neighbouring carbons (dehydrohalogenation), forming $\mathrm{C{=}C}$; the base $\mathrm{OH^{-}}$ removes $\mathrm{H^{+}}$.
(d) Addition across $\mathrm{C{=}O}$ started by the nucleophile $\mathrm{CN^{-}}$ (−E effect: the π electrons move to O) — nucleophilic addition.
(e) Substitution of H by Cl via chlorine free radicals formed by homolysis in light — free-radical substitution.
(f) Rearrangement (isomerisation): same formula $\mathrm{C_{4}H_{10}}$, new carbon skeleton.
Check Atom balance, e.g. (c): $\mathrm{C_{2}H_{5}Br+KOH}$ = $\mathrm{C_{2}H_{6}BrKO}$ = $\mathrm{C_{2}H_{4}+KBr+H_{2}O}$ ✓.

Exercises

I
Fission, intermediates and reagents.
  • 1 Define homolytic fission and state two conditions that favour it.
  • 2 Show, with a curved arrow, the heterolysis of the $\mathrm{C{-}Br}$ bond in $\mathrm{CH_{3}Br}$, and name the products.
  • 3 Give the hybridisation and shape of the carbon in $\mathrm{CH_{3}^{+}}$, $\mathrm{CH_{3}^{-}}$ and $\mathrm{CH_{3}^{\bullet}}$.
  • 4 How many valence electrons surround the reactive carbon of a carbocation, a carbanion and a free radical?
  • 5 Arrange $\mathrm{(CH_{3})_{3}C^{+}}$, $\mathrm{CH_{3}^{+}}$, $\mathrm{(CH_{3})_{2}CH^{+}}$, $\mathrm{CH_{3}CH_{2}^{+}}$ in order of decreasing stability.
  • 6 Arrange the free radicals $\mathrm{CH_{3}^{\bullet}}$, $\mathrm{(CH_{3})_{3}C^{\bullet}}$, $\mathrm{CH_{3}CH_{2}^{\bullet}}$ in order of increasing stability.
  • 7 Classify as electrophile or nucleophile: $\mathrm{Cl^{-}}$, $\mathrm{CH_{3}NH_{2}}$, $\mathrm{SO_{3}}$, $\mathrm{CH_{3}CH_{2}^{+}}$, $\mathrm{H_{2}O}$, $\mathrm{BF_{3}}$, $\mathrm{CN^{-}}$, $\mathrm{NO_{2}^{+}}$.
  • 8 In polar solvents $\mathrm{(CH_{3})_{3}C{-}Br}$ ionises far more easily than $\mathrm{CH_{3}{-}Br}$. Explain.
1 Each atom keeps one electron of the bond, giving free radicals. Favoured by heat or light (or peroxides), non-polar bonds, gas phase or non-polar solvents.
2 A full-headed arrow from the $\mathrm{C{-}Br}$ bond to Br: $\mathrm{CH_{3}{-}Br\rightarrow CH_{3}^{+}+Br^{-}}$ (methyl carbocation and bromide ion).
3 $\mathrm{CH_{3}^{+}}$: $sp^{2}$, trigonal planar. $\mathrm{CH_{3}^{-}}$: $sp^{3}$, pyramidal. $\mathrm{CH_{3}^{\bullet}}$: $sp^{2}$, planar (nearly).
4 6, 8 and 7.
5 $\mathrm{(CH_{3})_{3}C^{+}\gt(CH_{3})_{2}CH^{+}\gt CH_{3}CH_{2}^{+}\gt CH_{3}^{+}}$.
6 $\mathrm{CH_{3}^{\bullet}\lt CH_{3}CH_{2}^{\bullet}\lt(CH_{3})_{3}C^{\bullet}}$.
7 Nucleophiles: $\mathrm{Cl^{-}}$, $\mathrm{CH_{3}NH_{2}}$, $\mathrm{H_{2}O}$, $\mathrm{CN^{-}}$. Electrophiles: $\mathrm{SO_{3}}$, $\mathrm{CH_{3}CH_{2}^{+}}$, $\mathrm{BF_{3}}$, $\mathrm{NO_{2}^{+}}$.
8 Ionisation gives $\mathrm{(CH_{3})_{3}C^{+}}$, a 3° carbocation stabilised by three +I methyl groups and 9 hyperconjugating C–H bonds; $\mathrm{CH_{3}^{+}}$ has no such stabilisation.
II
Inductive and electromeric effects.
  • 1 Define the inductive effect. Is it permanent or temporary? Which electrons are displaced?
  • 2 Which is the stronger acid, $\mathrm{ClCH_{2}COOH}$ or $\mathrm{CH_{3}COOH}$? Why?
  • 3 Arrange $\mathrm{FCH_{2}COOH}$, $\mathrm{ICH_{2}COOH}$, $\mathrm{BrCH_{2}COOH}$, $\mathrm{ClCH_{2}COOH}$ in order of decreasing acid strength.
  • 4 Arrange $\mathrm{CH_{3}CHClCH_{2}COOH}$, $\mathrm{CH_{3}CH_{2}CHClCOOH}$ and $\mathrm{ClCH_{2}CH_{2}CH_{2}COOH}$ in order of decreasing acid strength.
  • 5 Which is a +I group: $\mathrm{{-}NO_{2}}$, $\mathrm{{-}CH_{3}}$, $\mathrm{{-}Cl}$ or $\mathrm{{-}CN}$?
  • 6 Define the electromeric effect and give two differences from the inductive effect.
  • 7 When $\mathrm{H^{+}}$ attacks propene, which effect operates, and in which direction do the π electrons move?
  • 8 Why does $\mathrm{Cl_{3}CCOOH}$ have a $\mathrm{p}K_{a}$ about 4 units lower than $\mathrm{CH_{3}COOH}$? By what factor is its $K_{a}$ larger? ($\mathrm{p}K_{a}$ 0.65 and 4.76.)
1 The permanent displacement of σ electrons along a chain caused by an atom or group of different electronegativity; it weakens with distance.
2 $\mathrm{ClCH_{2}COOH}$: the −I chlorine stabilises the carboxylate ion by spreading its negative charge.
3 $\mathrm{FCH_{2}COOH\gt ClCH_{2}COOH\gt BrCH_{2}COOH\gt ICH_{2}COOH}$.
4 2-chlorobutanoic ($\mathrm{CH_{3}CH_{2}CHClCOOH}$) $\gt$ 3-chlorobutanoic $\gt$ 4-chlorobutanoic: the closer the Cl to COOH, the stronger the effect.
5 $\mathrm{{-}CH_{3}}$.
6 The complete, temporary transfer of π electrons of a multiple bond to one atom in the presence of an attacking reagent. Differences: it is temporary (inductive is permanent); it involves π electrons (inductive: σ); it needs a reagent; it is a complete transfer, not a partial shift.
7 The +E effect: the π electrons move towards the carbon to which $\mathrm{H^{+}}$ becomes attached (C1), leaving the positive charge on C2, the 2° carbon.
8 Three −I chlorines stabilise $\mathrm{Cl_{3}CCOO^{-}}$ strongly. $\Delta\mathrm{p}K_{a}=4.11$, so $K_{a}$ is larger by $10^{4.11}\approx1.3\times10^{4}$.
III
Resonance and hyperconjugation.
  • 1 State three rules for writing resonance structures.
  • 2 Write the resonance structures of the methanoate ion $\mathrm{HCOO^{-}}$. What is the $\mathrm{C{-}O}$ bond order?
  • 3 Classify each group as +R or −R: $\mathrm{{-}OH}$, $\mathrm{{-}NH_{2}}$, $\mathrm{{-}NO_{2}}$, $\mathrm{{-}CHO}$, $\mathrm{{-}Cl}$, $\mathrm{{-}CN}$, $\mathrm{{-}OCH_{3}}$, $\mathrm{{-}COOH}$.
  • 4 Why are the ortho and para positions of aniline, $\mathrm{C_{6}H_{5}NH_{2}}$, electron-rich?
  • 5 How many hyperconjugative structures (α-hydrogens) do $\mathrm{(CH_{3})_{2}CH^{+}}$ and $\mathrm{CH_{3}CH_{2}CH_{2}^{+}}$ have?
  • 6 Arrange $\mathrm{CH_{2}{=}CH_{2}}$, $\mathrm{(CH_{3})_{2}C{=}C(CH_{3})_{2}}$, $\mathrm{CH_{3}CH{=}CH_{2}}$ and $\mathrm{CH_{3}CH{=}CHCH_{3}}$ in order of increasing stability.
  • 7 Why is the benzyl cation $\mathrm{C_{6}H_{5}CH_{2}^{+}}$ much more stable than the ethyl cation, although both are primary?
  • 8 Which pair are not resonance structures: (i) $\mathrm{CH_{2}{=}CH{-}\overset{+}{C}H_{2}}$ / $\mathrm{\overset{+}{C}H_{2}{-}CH{=}CH_{2}}$; (ii) $\mathrm{CH_{3}COCH_{3}}$ / $\mathrm{CH_{2}{=}C(OH)CH_{3}}$? Explain.
1 Any three of: same positions of nuclei; same number of unpaired electrons; only π electrons and lone pairs move; each structure a valid Lewis structure; more bonds, full octets and less charge separation give more important contributors.
2 $\mathrm{H{-}C({=}O){-}O^{-}\leftrightarrow H{-}C({-}O^{-}){=}O}$; two equivalent structures, so each $\mathrm{C{-}O}$ bond has bond order $\tfrac{1+2}{2}=1.5$.
3 +R: $\mathrm{{-}OH}$, $\mathrm{{-}NH_{2}}$, $\mathrm{{-}Cl}$, $\mathrm{{-}OCH_{3}}$. −R: $\mathrm{{-}NO_{2}}$, $\mathrm{{-}CHO}$, $\mathrm{{-}CN}$, $\mathrm{{-}COOH}$.
4 The lone pair on N is delocalised into the ring (+R); the resonance structures place negative charge on C2, C4 and C6.
5 $\mathrm{(CH_{3})_{2}CH^{+}}$: 6. $\mathrm{CH_{3}CH_{2}CH_{2}^{+}}$: only the $\mathrm{CH_{2}}$ next to the charged carbon counts → 2.
6 α-H 0, 3, 6, 12: $\mathrm{CH_{2}{=}CH_{2}\lt CH_{3}CH{=}CH_{2}\lt CH_{3}CH{=}CHCH_{3}\lt(CH_{3})_{2}C{=}C(CH_{3})_{2}}$.
7 The positive charge is delocalised by resonance over the $\mathrm{CH_{2}}$ carbon and the ortho and para ring carbons (four contributing structures with the charge on carbon); ethyl has only weak hyperconjugation (3 α-H).
8 (ii): an H atom has moved from C to O, so these are two different compounds (keto and enol tautomers). Pair (i) differs only in π electrons — true resonance structures.
IV
Types of reactions.
  • 1 Classify: $\mathrm{CH_{3}I+OH^{-}\rightarrow CH_{3}OH+I^{-}}$.
  • 2 Classify: $\mathrm{CH_{3}CH{=}CH_{2}+H_{2}\xrightarrow{Ni}CH_{3}CH_{2}CH_{3}}$.
  • 3 Classify: $\mathrm{(CH_{3})_{3}CBr+KOH\ (alc.)\rightarrow(CH_{3})_{2}C{=}CH_{2}+KBr+H_{2}O}$.
  • 4 Classify: $\mathrm{CH_{4}+Br_{2}\xrightarrow{h\nu}CH_{3}Br+HBr}$, and name the species that attacks methane.
  • 5 Classify: hexane → 2-methylpentane, catalysed by $\mathrm{AlCl_{3}}$ and HCl.
  • 6 In $\mathrm{CH_{3}COCH_{3}+HCN\rightarrow(CH_{3})_{2}C(OH)CN}$, which atom does the nucleophile attack, and why?
  • 7 In reaction 1, identify the substrate, the reagent and the leaving group.
  • 8 Applied. In the stratosphere, ultraviolet light breaks a $\mathrm{C{-}Cl}$ bond of the refrigerant $\mathrm{CF_{2}Cl_{2}}$. What kind of fission is this, what species are formed, and why do they destroy ozone so efficiently?
1 Nucleophilic substitution ($\mathrm{OH^{-}}$ replaces $\mathrm{I^{-}}$).
2 Addition (hydrogenation) across $\mathrm{C{=}C}$.
3 Elimination (dehydrohalogenation) giving 2-methylpropene.
4 Free-radical substitution; the attacking species is the bromine radical $\mathrm{Br^{\bullet}}$ formed by homolysis of $\mathrm{Br_{2}}$.
5 Rearrangement (isomerisation).
6 The carbonyl carbon: O is more electronegative, so the $\mathrm{C{=}O}$ carbon is $\delta+$ (and the π electrons can move onto O, −E effect). $\mathrm{CN^{-}}$ attacks it — nucleophilic addition.
7 Substrate $\mathrm{CH_{3}I}$; reagent (nucleophile) $\mathrm{OH^{-}}$; leaving group $\mathrm{I^{-}}$.
8 Homolysis: $\mathrm{CF_{2}Cl_{2}\xrightarrow{h\nu}{}^{\bullet}CF_{2}Cl+Cl^{\bullet}}$. The chlorine radical reacts with $\mathrm{O_{3}}$ to give $\mathrm{ClO^{\bullet}}$ and $\mathrm{O_{2}}$, and $\mathrm{ClO^{\bullet}}$ is turned back into $\mathrm{Cl^{\bullet}}$ by an O atom: a chain reaction in which one Cl radical destroys thousands of ozone molecules.
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. NEET The most stable carbocation is (A) $\mathrm{CH_{3}^{+}}$ (B) $\mathrm{CH_{3}CH_{2}^{+}}$ (C) $\mathrm{(CH_{3})_{2}CH^{+}}$ (D) $\mathrm{(CH_{3})_{3}C^{+}}$

Idea Most +I alkyl groups and most α-hydrogens (9): the tertiary ion.
Answer (D).

E2. JEE Main The hybridisation of carbon and the shape of the methyl carbanion $\mathrm{CH_{3}^{-}}$ are (A) $sp^{2}$, trigonal planar (B) $sp^{3}$, pyramidal (C) $sp^{3}$, tetrahedral with four bond pairs (D) $sp$, linear

Idea Three bond pairs + one lone pair = four electron pairs → $sp^{3}$; with one position taken by the lone pair the shape is pyramidal (like $\mathrm{NH_{3}}$).
Answer (B).

E3. JEE Main Numerical answer. How many hyperconjugative structures (involving C–H bonds) can be written for the 2-methylbutan-2-yl cation, $\mathrm{(CH_{3})_{2}\overset{+}{C}CH_{2}CH_{3}}$?

Count α-carbons: two $\mathrm{CH_{3}}$ (3 H each) and one $\mathrm{CH_{2}}$ (2 H). The $\mathrm{CH_{3}}$ of the ethyl group is a β-carbon and does not count.
Answer $3+3+2=$ 8.

E4. Board · 3 marks Distinguish between the inductive effect and the electromeric effect (three points), with one example of each.

1 mark Inductive: permanent; electromeric: temporary, only while the reagent is present.
1 mark Inductive: partial shift of σ electrons along a chain; electromeric: complete transfer of π electrons of a multiple bond.
1 mark Examples: $\mathrm{CH_{3}CH_{2}Cl}$ (Cl pulls σ electrons, −I); $\mathrm{H^{+}}$ attacking $\mathrm{CH_{2}{=}CH_{2}}$ (π electrons move to the carbon attacked, +E).

E5. NEET Which of the following is a nucleophile? (A) $\mathrm{BF_{3}}$ (B) $\mathrm{NH_{3}}$ (C) $\mathrm{NO_{2}^{+}}$ (D) $\mathrm{AlCl_{3}}$

Idea A nucleophile donates a pair: $\mathrm{NH_{3}}$ has a lone pair on N. $\mathrm{BF_{3}}$ and $\mathrm{AlCl_{3}}$ (six electrons on the central atom) and $\mathrm{NO_{2}^{+}}$ are electrophiles.
Answer (B).

E6. JEE Advanced One or more options correct. (A) A carbocation carbon is $sp^{2}$ hybridised with a vacant $p$ orbital. (B) Alkyl free radicals are paramagnetic. (C) The $\mathrm{{-}NO_{2}}$ group shows a +R effect. (D) Hyperconjugation involves the σ electrons of C–H bonds on the carbon adjacent to the electron-deficient centre.

(A), (B) True: three σ bonds, planar, empty $p$. True: one unpaired electron.
(C) False: $\mathrm{{-}NO_{2}}$ withdraws π electrons (−R).
(D) True: that is the definition (α-C–H bonds).
Answer (A), (B), (D).
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. Chlorofluorocarbons (CFCs) such as $\mathrm{CF_{2}Cl_{2}}$ are so unreactive that they were once used in fridges, aerosol cans and asthma inhalers. Why, then, do they destroy ozone 20–40 km above our heads — and why is it the $\mathrm{C{-}Cl}$ bond, not the $\mathrm{C{-}F}$ bond, that breaks?

Why it matters. The discovery that CFCs release chlorine free radicals in the stratosphere earned Mario Molina, Sherwood Rowland and Paul Crutzen the 1995 Nobel Prize in Chemistry. The Montreal Protocol (1987), signed by every member state of the United Nations, phased CFCs out; the ozone layer is now expected to recover by about the middle of this century. It is one of the clearest cases of mechanistic chemistry changing world policy.

What to do.

  1. Write the homolysis of one $\mathrm{C{-}Cl}$ bond of $\mathrm{CF_{2}Cl_{2}}$ with fish-hook arrows, and the two chain-carrying steps $\mathrm{Cl^{\bullet}+O_{3}\rightarrow ClO^{\bullet}+O_{2}}$ and $\mathrm{ClO^{\bullet}+O\rightarrow Cl^{\bullet}+O_{2}}$. Add them: what is the net reaction, and what role does $\mathrm{Cl^{\bullet}}$ play?
  2. The bond enthalpies are about $320\ \mathrm{kJ\,mol^{-1}}$ for $\mathrm{C{-}Cl}$ and $485\ \mathrm{kJ\,mol^{-1}}$ for $\mathrm{C{-}F}$ in CFCs. Use $E=h\nu=hc/\lambda$ per molecule (Section 2.2) to find the longest wavelength of light able to break each bond.
  3. Which bond breaks first, and why are the fluorine atoms "safe"?
  4. Find out what replaced CFCs (HCFCs, then HFCs and hydrocarbons such as isobutane, R-600a, in fridges) and why HFCs are now themselves being phased down under the Kigali Amendment (2016).

Tools: Definition 8.8 and Interactive 1, $h=6.626\times10^{-34}\ \mathrm{J\,s}$, $c=3.0\times10^{8}\ \mathrm{m\,s^{-1}}$, $N_{A}=6.022\times10^{23}\ \mathrm{mol^{-1}}$, a UN Environment Programme ozone fact sheet.

Hand in: The mechanism with curved arrows, your two wavelength calculations, and a short paragraph on the replacements.

Going further. Your wavelength for the $\mathrm{C{-}Cl}$ bond lies in the near ultraviolet, which does reach the lower atmosphere — yet CFCs survive there for decades. What else, besides bond energy, decides whether light can break a bond?

Instructor note: net $\mathrm{O_{3}+O\rightarrow2O_{2}}$; $\mathrm{Cl^{\bullet}}$ is a catalyst (regenerated). $\lambda=hcN_{A}/E$: $0.1197\ \mathrm{J\,m\,mol^{-1}}/3.20\times10^{5}=3.74\times10^{-7}\ \mathrm{m}$ (374 nm) for C–Cl; $2.47\times10^{-7}\ \mathrm{m}$ (247 nm) for C–F. C–Cl breaks first. Going further: the molecule must also absorb the light; CFCs absorb only below about 220 nm, radiation that is filtered out before it reaches the lower atmosphere, so photolysis happens only high in the stratosphere.

🧪 Self-check quiz

0/5 correct

Q1. Homolytic fission of a covalent bond produces

Q2. The positively charged carbon of a carbocation is

Q3. The strongest acid is

Q4. Which group shows a −R effect?

Q5. The number of α-hydrogens available for hyperconjugation in $\mathrm{(CH_{3})_{2}CH^{+}}$ is

← 8.2 IUPAC Nomenclature and Isomerism8.4 Purification, Qualitative and Quantitative Analysis →